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Salsk061 [2.6K]
4 years ago
14

Can you please answer asap please thanks is it A B C D

Mathematics
1 answer:
Keith_Richards [23]4 years ago
7 0
A cylinder can be made from this net.

I hope this helps :)
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5/9 × 8/11 what is the answer
nikitadnepr [17]

Answer:

\frac{5}{9}  \times  \frac{8}{11}  =  \frac{5 \times 8}{9 \times 11}  =  \frac{40}{99}  = 0.4040

7 0
3 years ago
Read 2 more answers
Determine if the graph is symmetric about the x-axis, the y-axis, or the origin. r = 5 cos 3θ
Aneli [31]
First of all get a picture of what this looks like. Desmos is a pretty good graphing program, but anything that will do polar coordinates will work. 

Here is the graph.

What you can see is that this graph is symmetrical around the x axis.

When you talk about symmetry, you can think of it as taking a mirror and putting it where you think there is symmetry. If you can't tell the difference between the image and the real thing, then you have symmetry.

In this case, the mirror will show symmetry along the x axis and no where else.

4 0
3 years ago
Read 2 more answers
How do you make 'a' the subject of the formula: ab - cd = ac
xxTIMURxx [149]
First bring all terms in 'a' to the left side of the formula by subtracting ac from both sides

ab - ac - cd = ac - ac 
ab - ac - cd = 0
now add cd to both sides
ab - ac -cd + cd = cd
ab - ac = cd
now factor the left side by taking out the 'a' 
a(b-c) = cd
now divide both sides by (b-c)
a = cd / (b-c)
done

3 0
4 years ago
A ball is thrown from an initial height of 1 meter with an initial upward velocity of 15m/s. The ball's height h (in meters) aft
lakkis [162]

\bf \stackrel{\textit{ball's height}~\hfill }{\stackrel{\downarrow }{h}=1+15t-5t^2}\implies \stackrel{\textit{ball's height}~\hfill }{\stackrel{\downarrow }{6}=1+15t-5t^2}\implies 0=-5+15t-5t^2 \\\\\\ ~~~~~~~~~~~~\textit{using the quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{5}t^2\stackrel{\stackrel{b}{\downarrow }}{-15}t\stackrel{\stackrel{c}{\downarrow }}{+5}=0 \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}

\bf t=\cfrac{-(-15)\pm\sqrt{(-15)^2-4(5)(5)}}{2(5)}\implies t=\cfrac{15\pm\sqrt{225-100}}{10} \\\\\\ t=\cfrac{15\pm\sqrt{125}}{10}\implies t=\cfrac{15\pm\sqrt{5^2 \cdot 5}}{10}\implies t=\cfrac{\stackrel{3}{~~\begin{matrix} 15 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\pm ~~\begin{matrix} 5 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~\sqrt{5}}{\underset{2}{~~\begin{matrix} 10 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}

\bf t=\cfrac{3\pm \sqrt{5}}{2}\implies t= \begin{cases} \frac{3+ \sqrt{5}}{2} \approx 2.618\\\\ \frac{3- \sqrt{5}}{2}\approx 0.382 \end{cases}

8 0
3 years ago
What is the square root of 9
GalinKa [24]
3 is the answer. 3 X 3 = 9
I'm glad I could help
7 0
4 years ago
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